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प्रश्न
Prove that: `(sin A + cos A)/(sin A - cos A) + (sin A - cos A)/(sin A + cos A) = 2/(sin^2 A - cos^2 A)`.
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उत्तर
LHS = `(sin A + cos A)/(sin A - cos A) + (sin A - cos A)/(sin A + cos A)`
= `((sin A + cos A)^2 + (sin A - cos A)^2)/((sin A - cos A)(sin A + cos A))`
= `(sin^2 A + cos^2 A + 2 sin A cos A + sin^2 A + cos^2 A - 2sin A. cos A)/(sin^2 A - cos^2 A)`
= `2(sin^2 A + cos^2 A)/(sin^A - cos^2 A)`
= `2/(sin^2 A - cos^2 A)`
= RHS
Hence proved.
संबंधित प्रश्न
Prove the following trigonometric identities.
`sin theta/(1 - cos theta) = cosec theta + cot theta`
Prove the following trigonometric identities
cosec6θ = cot6θ + 3 cot2θ cosec2θ + 1
Prove the following identities:
(sec A – cos A) (sec A + cos A) = sin2 A + tan2 A
If 4 cos2 A – 3 = 0, show that: cos 3 A = 4 cos3 A – 3 cos A
`(1+ tan theta + cot theta )(sintheta - cos theta) = ((sec theta)/ (cosec^2 theta)-( cosec theta)/(sec^2 theta))`
If tan A = n tan B and sin A = m sin B , prove that `cos^2 A = ((m^2-1))/((n^2 - 1))`
Prove the following identity :
`(secA - 1)/(secA + 1) = (1 - cosA)/(1 + cosA)`
Prove that `"cosec" θ xx sqrt(1 - cos^2θ) = 1`.
Show that, cotθ + tanθ = cosecθ × secθ
Solution :
L.H.S. = cotθ + tanθ
= `cosθ/sinθ + sinθ/cosθ`
= `(square + square)/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ............... `square`
= `1/sinθ xx 1/square`
= cosecθ × secθ
L.H.S. = R.H.S
∴ cotθ + tanθ = cosecθ × secθ
(sec2 θ – 1) (cosec2 θ – 1) is equal to ______.
